# EMI testing with 10 V/M immunity

Discussion in 'Electronic Design' started by [email protected], Mar 23, 2007.

1. ### Guest

Hi Group,

I bit of confusion reviewing the E(V/M) equation for power gives:

P(watts) = (E^2*r^2)/(30xantenna gain), at 10V per meter this is 14.64
Watts of input power.

The equation, out of the EMC textbook, does not include frequency.

If I put 14.64 watts into a dipole at 100 MHz and measure the V/M at
three meters and then repeated this with the frequency moved to 300
MHz then both would read the same V/M?

I thought that the higher the frequency, more of the energy is lost so
an equal input power at 100 MHz will have less space loss than an
equal input at 300 MHz. This will result in a higher level of
interference to the device under test at 3 meters.

pdrunen

2. ### Fred BloggsGuest

The relationship between source power and field strength for a wave
propagating in free space is derived from the Poynting vector to give
S=|E|^2/Zo W/m^2 where Zo is the impedance of free space ~377 ohms.
Power density at a distance R, due to source power Po, is also
|S|=Po*G/(4pi*R^2) where G is the antenna gain, so that the E field
strength is computed from |E|=sqrt(|S|*Zo). This makes
Po=(4pi*R^2)/G*|S|=((4pi*R^2)/G)*|E|^2/Zo.
Your thinking is not right. There are two propagation loss factors: 1)
spherical spreading and 2) atmospheric absorption, only spherical
spreading is significant at short distances and you already have the
equation for that. You are missing out on a fundamental relationship
between antenna peak gain,G, and effective aperture area, A, which due
to arguments relating to reciprocity of xmit and receive can be shown to
have a constant ratio of G/A=4pi/lamda^2 where lamda is the wavelength.
Since you are effectively maintaining constant A while increasing the
frequency by a factor of x3, the peak gain at 300MHz will be 9x that at
100MHz. Therefore your |S| will increase by x9 due to G, and your |E|
will increase by sqrt(|S|) or by x3, for the the same Po.